Answer :
Certainly! Let's solve each system of equations using the elimination method:
### (a) [tex]\( x + 2y = 8 \)[/tex] and [tex]\( 2x + 3y = 13 \)[/tex]
1. Given the equations:
[tex]\[ \begin{cases} x + 2y = 8 \\ 2x + 3y = 13 \end{cases} \][/tex]
2. Multiply the first equation by 2 to make the coefficients of [tex]\( x \)[/tex] the same:
[tex]\[ \begin{cases} 2(x + 2y) = 2 \cdot 8 \\ 2x + 3y = 13 \end{cases} \][/tex]
This simplifies to:
[tex]\[ \begin{cases} 2x + 4y = 16 \\ 2x + 3y = 13 \end{cases} \][/tex]
3. Subtract the second equation from the first to eliminate [tex]\( x \)[/tex]:
[tex]\[ (2x + 4y) - (2x + 3y) = 16 - 13 \][/tex]
Simplifying, we get:
[tex]\[ y = 3 \][/tex]
4. Substitute [tex]\( y = 3 \)[/tex] back into the first equation:
[tex]\[ x + 2(3) = 8 \][/tex]
Simplifying, we get:
[tex]\[ x + 6 = 8 \implies x = 2 \][/tex]
So, the solution for (a) is [tex]\( x = 2 \)[/tex] and [tex]\( y = 3 \)[/tex].
### (b) [tex]\( 3x - 2y = 7 \)[/tex] and [tex]\( 5x + y = 3 \)[/tex]
1. Given the equations:
[tex]\[ \begin{cases} 3x - 2y = 7 \\ 5x + y = 3 \end{cases} \][/tex]
2. Multiply the second equation by 2 to align the coefficients of [tex]\( y \)[/tex]:
[tex]\[ \begin{cases} 3x - 2y = 7 \\ 2(5x + y) = 2 \cdot 3 \end{cases} \][/tex]
This simplifies to:
[tex]\[ \begin{cases} 3x - 2y = 7 \\ 10x + 2y = 6 \end{cases} \][/tex]
3. Add the two equations to eliminate [tex]\( y \)[/tex]:
[tex]\[ (3x - 2y) + (10x + 2y) = 7 + 6 \][/tex]
Simplifying, we get:
[tex]\[ 13x = 13 \implies x = 1 \][/tex]
4. Substitute [tex]\( x = 1 \)[/tex] back into the second equation:
[tex]\[ 5(1) + y = 3 \][/tex]
Simplifying, we get:
[tex]\[ 5 + y = 3 \implies y = -2 \][/tex]
So, the solution for (b) is [tex]\( x = 1 \)[/tex] and [tex]\( y = -2 \)[/tex].
### (c) [tex]\( 3x - y = 4 \)[/tex] and [tex]\( x + 2y = 13 \)[/tex]
1. Given the equations:
[tex]\[ \begin{cases} 3x - y = 4 \\ x + 2y = 13 \end{cases} \][/tex]
2. Multiply the second equation by 3 to align the coefficients of [tex]\( x \)[/tex]:
[tex]\[ \begin{cases} 3x - y = 4 \\ 3(x + 2y) = 3 \cdot 13 \end{cases} \][/tex]
This simplifies to:
[tex]\[ \begin{cases} 3x - y = 4 \\ 3x + 6y = 39 \end{cases} \][/tex]
3. Subtract the first equation from the second to eliminate [tex]\( x \)[/tex]:
[tex]\[ (3x + 6y) - (3x - y) = 39 - 4 \][/tex]
Simplifying, we get:
[tex]\[ 7y = 35 \implies y = 5 \][/tex]
4. Substitute [tex]\( y = 5 \)[/tex] back into the second equation:
[tex]\[ x + 2(5) = 13 \][/tex]
Simplifying, we get:
[tex]\[ x + 10 = 13 \implies x = 3 \][/tex]
So, the solution for (c) is [tex]\( x = 3 \)[/tex] and [tex]\( y = 5 \)[/tex].
### (d) [tex]\( 3x + 4y = 1 \)[/tex] and [tex]\( 3x + 2y = 1 \)[/tex]
1. Given the equations:
[tex]\[ \begin{cases} 3x + 4y = 1 \\ 3x + 2y = 1 \end{cases} \][/tex]
2. Subtract the second equation from the first to eliminate [tex]\( x \)[/tex]:
[tex]\[ (3x + 4y) - (3x + 2y) = 1 - 1 \][/tex]
Simplifying, we get:
[tex]\[ 2y = 0 \implies y = 0 \][/tex]
3. Substitute [tex]\( y = 0 \)[/tex] back into the first equation:
[tex]\[ 3x + 4(0) = 1 \][/tex]
Simplifying, we get:
[tex]\[ 3x = 1 \implies x = \frac{1}{3} \][/tex]
So, the solution for (d) is [tex]\( x = \frac{1}{3} \)[/tex] and [tex]\( y = 0 \)[/tex].
### (a) [tex]\( x + 2y = 8 \)[/tex] and [tex]\( 2x + 3y = 13 \)[/tex]
1. Given the equations:
[tex]\[ \begin{cases} x + 2y = 8 \\ 2x + 3y = 13 \end{cases} \][/tex]
2. Multiply the first equation by 2 to make the coefficients of [tex]\( x \)[/tex] the same:
[tex]\[ \begin{cases} 2(x + 2y) = 2 \cdot 8 \\ 2x + 3y = 13 \end{cases} \][/tex]
This simplifies to:
[tex]\[ \begin{cases} 2x + 4y = 16 \\ 2x + 3y = 13 \end{cases} \][/tex]
3. Subtract the second equation from the first to eliminate [tex]\( x \)[/tex]:
[tex]\[ (2x + 4y) - (2x + 3y) = 16 - 13 \][/tex]
Simplifying, we get:
[tex]\[ y = 3 \][/tex]
4. Substitute [tex]\( y = 3 \)[/tex] back into the first equation:
[tex]\[ x + 2(3) = 8 \][/tex]
Simplifying, we get:
[tex]\[ x + 6 = 8 \implies x = 2 \][/tex]
So, the solution for (a) is [tex]\( x = 2 \)[/tex] and [tex]\( y = 3 \)[/tex].
### (b) [tex]\( 3x - 2y = 7 \)[/tex] and [tex]\( 5x + y = 3 \)[/tex]
1. Given the equations:
[tex]\[ \begin{cases} 3x - 2y = 7 \\ 5x + y = 3 \end{cases} \][/tex]
2. Multiply the second equation by 2 to align the coefficients of [tex]\( y \)[/tex]:
[tex]\[ \begin{cases} 3x - 2y = 7 \\ 2(5x + y) = 2 \cdot 3 \end{cases} \][/tex]
This simplifies to:
[tex]\[ \begin{cases} 3x - 2y = 7 \\ 10x + 2y = 6 \end{cases} \][/tex]
3. Add the two equations to eliminate [tex]\( y \)[/tex]:
[tex]\[ (3x - 2y) + (10x + 2y) = 7 + 6 \][/tex]
Simplifying, we get:
[tex]\[ 13x = 13 \implies x = 1 \][/tex]
4. Substitute [tex]\( x = 1 \)[/tex] back into the second equation:
[tex]\[ 5(1) + y = 3 \][/tex]
Simplifying, we get:
[tex]\[ 5 + y = 3 \implies y = -2 \][/tex]
So, the solution for (b) is [tex]\( x = 1 \)[/tex] and [tex]\( y = -2 \)[/tex].
### (c) [tex]\( 3x - y = 4 \)[/tex] and [tex]\( x + 2y = 13 \)[/tex]
1. Given the equations:
[tex]\[ \begin{cases} 3x - y = 4 \\ x + 2y = 13 \end{cases} \][/tex]
2. Multiply the second equation by 3 to align the coefficients of [tex]\( x \)[/tex]:
[tex]\[ \begin{cases} 3x - y = 4 \\ 3(x + 2y) = 3 \cdot 13 \end{cases} \][/tex]
This simplifies to:
[tex]\[ \begin{cases} 3x - y = 4 \\ 3x + 6y = 39 \end{cases} \][/tex]
3. Subtract the first equation from the second to eliminate [tex]\( x \)[/tex]:
[tex]\[ (3x + 6y) - (3x - y) = 39 - 4 \][/tex]
Simplifying, we get:
[tex]\[ 7y = 35 \implies y = 5 \][/tex]
4. Substitute [tex]\( y = 5 \)[/tex] back into the second equation:
[tex]\[ x + 2(5) = 13 \][/tex]
Simplifying, we get:
[tex]\[ x + 10 = 13 \implies x = 3 \][/tex]
So, the solution for (c) is [tex]\( x = 3 \)[/tex] and [tex]\( y = 5 \)[/tex].
### (d) [tex]\( 3x + 4y = 1 \)[/tex] and [tex]\( 3x + 2y = 1 \)[/tex]
1. Given the equations:
[tex]\[ \begin{cases} 3x + 4y = 1 \\ 3x + 2y = 1 \end{cases} \][/tex]
2. Subtract the second equation from the first to eliminate [tex]\( x \)[/tex]:
[tex]\[ (3x + 4y) - (3x + 2y) = 1 - 1 \][/tex]
Simplifying, we get:
[tex]\[ 2y = 0 \implies y = 0 \][/tex]
3. Substitute [tex]\( y = 0 \)[/tex] back into the first equation:
[tex]\[ 3x + 4(0) = 1 \][/tex]
Simplifying, we get:
[tex]\[ 3x = 1 \implies x = \frac{1}{3} \][/tex]
So, the solution for (d) is [tex]\( x = \frac{1}{3} \)[/tex] and [tex]\( y = 0 \)[/tex].
